$a \Rightarrow b$, $b \Rightarrow c$, $c \Rightarrow d$, $d \Rightarrow a$. Argue that any two of these statements are logically equivalent. Suppose a,b,c and d are statements such that $a \Rightarrow b$, $b \Rightarrow c$, $c \Rightarrow d$, $d \Rightarrow a$.  Argue that any two of these statements are logically equivalent.
Hey,  Im confused as to how to argue that this is true.  Is it similar to how you would prove a hypothetical syllogism?  Unfortunately this has me stumped and I can't even figure out how to start off.
 A: Just write this chain two times: $a \Rightarrow b \Rightarrow c \Rightarrow d \Rightarrow a \Rightarrow b \Rightarrow c \Rightarrow d \Rightarrow a$
(this is true, it follows from what you have)    
Now from here it's obvious that $x \Rightarrow y$  for any $x,y\in\{a,b,c,d\}$  
A: This is a circular graph with the statements $x_i$ as vertices and the implications $x_i \implies x_j$ as directed edges $(x_i, x_j)$.
This means you can reach any two statements $x$, $y$ via a finite number of implication, or $x \implies y$, because of the transitivity of the implication: If $x \implies y$ and  $y \implies z$, then $x \implies z$.
As $x \iff y$ means that the two implications $x \implies y$ and $y \implies x$ hold, every pair of statments $x$ and $y$ is equivalent, thus all are equivalent. 
A: Implication is a transitive relation. A transitive relation is a relation such that if $a \sim b$ and $b \sim c$, then $a \sim c$. So if $a \Rightarrow b$ and $b \Rightarrow c$ then $a \Rightarrow c$. Thus $a \Rightarrow c$ and $c \Rightarrow d$ imply that $a \Rightarrow d$ and $a \Rightarrow d$ and $d \Rightarrow a$ imply $a \Rightarrow a$. 
